JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 5)
An object of mass m is suspended at the end of a massless wire of length L and area of crosssection A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its
frequency of oscillation along the vertical direction is :
$$f = {1 \over {2\pi }}\sqrt {{{YA} \over {mL}}} $$
$$f = {1 \over {2\pi }}\sqrt {{{mL} \over {YA}}} $$
$$f = {1 \over {2\pi }}\sqrt {{{YL} \over {mA}}} $$
$$f = {1 \over {2\pi }}\sqrt {{{mA} \over {YL}}} $$
توضیح
An elastic wire can be treated as a spring with
k = $${{YA} \over l}$$
T = $$2\pi \sqrt {{m \over k}} $$
$$ \Rightarrow $$ f = $${1 \over {2\pi }}\sqrt {{k \over m}} $$ = $${1 \over {2\pi }}\sqrt {{{YA} \over {ml}}} $$
k = $${{YA} \over l}$$
T = $$2\pi \sqrt {{m \over k}} $$
$$ \Rightarrow $$ f = $${1 \over {2\pi }}\sqrt {{k \over m}} $$ = $${1 \over {2\pi }}\sqrt {{{YA} \over {ml}}} $$